### Author Topic: Thanksgiving bonus 4  (Read 1389 times)

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Thanksgiving bonus 4
« on: October 05, 2018, 05:49:09 PM »
Lagrange equation is of the form

y= x\varphi (y')+\psi (y')
\label{eq1}

with $\varphi(p)-p\ne 0$. To solve it we plug $p=y'$ and differentiate equation:

pdx= \varphi(p)dx + \bigl(x\varphi'(p) +\psi'(p)\bigr)dp.
\label{eq2}

This is a linear ODE with respect to $x$. We find the general solution $x=f(p,C)$ and then $y=f(p,c)\varphi(p)+\psi(p)$:

\left\{\begin{aligned}
&x=f(p,C)\\
&y=f(p,c)\varphi(p)+\psi(p)
\end{aligned}\right.
\label{eq4}

gives us a solution in the parametric form.

(\ref{eq1}) can have a singular solution (or solutions)

y=x\varphi(c)+\psi(c),
\label{eq5}

where $c$ is a root of equation $\varphi(c)-c=0$.

Problem.
Find general and singular solutions to
$$y= 2xy'-3(y')^2.$$
« Last Edit: October 05, 2018, 06:01:49 PM by Victor Ivrii »

#### Monika Dydynski

• Full Member
• Posts: 26
• Karma: 30
##### Re: Thanksgiving bonus 4
« Reply #1 on: October 06, 2018, 12:22:25 AM »
Find general and special solutions to $$y= 2xy'-3(y')^2.$$

Substituting $y'=p$, we get an equation of the form $y=2xp–3{p^2}$

Differentiating both sides, we get

$dy=2xdp+2pdx-6pdp$

$dy=pdx$   $\Rightarrow$   $pdx=2xdp+2pdx-6pdp$    $\Rightarrow$    $–pdx=2xdp–6pdp$

Dividing by $p$, we get $-dx={2x \over p}dp-6dp$   $\Rightarrow$   ${dx \over dp} + {2 \over p}x-6=0 \tag{1}$

We get a linear differential equation. To solve $(1)$, we use an integrating factor given by $\mu=e^{\int{2 \over p}dp}=e^{2\ln|p|}=p^2$

Rewrite $(1)$ as ${2 \over p}x+{dx \over dp}=6$ and multiply by the integrating factor, $\mu(p)=p^2$ to get,

$$p^2{2 \over p}x+p^2{dx \over dp}=p^2 6$$
$${d \over dp}(p^2 x)=p^2 6$$
$$x={{2p^3+C} \over p^2}$$

Thus, the general solution to (1) is
$$x={{2p + {C \over p^2}}}\tag{2}$$

Substituting $(2)$ into the Lagrange Equation, we get

$$y={2\left({2p+\frac{C}{{{p^2}}}}\right)p–3p^2}=p^2+ \frac{2C}p$$

Thus,

\left\{\begin{aligned} &x={{2p + {C \over p^2}}}\\ &y=p^2+ \frac{2C}p \end{aligned}\right. \tag{3}

gives us a solution in the parametric form.

The Lagrange equation, $y= 2xy'-3(y')^2$, can also have a special solution (or solutions)

$\varphi (p)-p=0$

$2p–p=0$   $\Rightarrow$    $p(2-p)=0$    $\Rightarrow$    $p_1=0$ and $p_2=2$

Thus, the special solutions are
$$y=x\varphi (0)+\psi (0)=x \cdot 0+0=0.$$
« Last Edit: October 06, 2018, 01:05:51 AM by Monika Dydynski »

#### Zhuojing Yu

• Newbie
• Posts: 3
• Karma: 2
##### Re: Thanksgiving bonus 4
« Reply #2 on: October 06, 2018, 12:56:24 AM »
Here is the solution